Question

Question 5. The following table summarizes the results of the test on the three branches.

SUMMARY
Groups	Count	Sum	Average	Variance
Branch 1	5	266	53.2	177.2
Branch 2	5	340	68	147.5
Branch 3	5	389	77.8	178.7
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1533.73	2.00	766.87	4.57	0.03	3.89
Within Groups	2013.60	12.00	167.80
Total	3547.33	14.00

a) Interpret the results of the ANOVA.

b) Which branch deserves the reward for its performance?

Answer 1

Ans a)

To interpret the results you need to compare the F values given in the data. We can see that the calculated value of F (given as 4.57) is more than the critical value (given as 3.89). Hence, we'd reject the null hypothesis (which going by the standard objective and application of ANOVA would be that the means of the given branches are equal). Our conclusion is that the given branches are very different from each other - or (taking a hint about what's being studied from option b) of your question) the average performances across all branches vary significantly.

Ans b)

If you want to reward performance based on who achieved the highest test results - then the mean values (given in the data) are sufficient to make a decision. Branch 3, having the highest average, seems to be the best of the lot.

However, if you wish to go deeper and study the variability across the given branches, you could compute the coefficient of variation and compare the results.

Branch	Average	Variance	Standard Deviation (s.d.)= √variance	Coefficient of Variation
Branch 1	53.2	177.2	13.31	25.01
Branch 2	68	147.5	12.14	17.85
Branch 3	77.8	178.7	13.36	17.17

Note*: Coefficient of Variation = 100 x (s.d./Mean)                [Where s.d. = √variance]

We can see that- Branch 3 has the lowest coefficient of variation - meaning that performance in Branch 3 is the most consistent of the lot.

Both ways, Branch 3 seems to deserve the performance reward the most.